Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $z \neq 0$. $r = \dfrac{z - 2}{z^2 - 6z - 16} \times \dfrac{z + 2}{7z - 14} $
Answer: First factor the quadratic. $r = \dfrac{z - 2}{(z + 2)(z - 8)} \times \dfrac{z + 2}{7z - 14} $ Then factor out any other terms. $r = \dfrac{z - 2}{(z + 2)(z - 8)} \times \dfrac{z + 2}{7(z - 2)} $ Then multiply the two numerators and multiply the two denominators. $r = \dfrac{ (z - 2) \times (z + 2) } { (z + 2)(z - 8) \times 7(z - 2) } $ $r = \dfrac{ (z - 2)(z + 2)}{ 7(z + 2)(z - 8)(z - 2)} $ Notice that $(z - 2)$ and $(z + 2)$ appear in both the numerator and denominator so we can cancel them. $r = \dfrac{ (z - 2)\cancel{(z + 2)}}{ 7\cancel{(z + 2)}(z - 8)(z - 2)} $ We are dividing by $z + 2$ , so $z + 2 \neq 0$ Therefore, $z \neq -2$ $r = \dfrac{ \cancel{(z - 2)}\cancel{(z + 2)}}{ 7\cancel{(z + 2)}(z - 8)\cancel{(z - 2)}} $ We are dividing by $z - 2$ , so $z - 2 \neq 0$ Therefore, $z \neq 2$ $r = \dfrac{1}{7(z - 8)} ; \space z \neq -2 ; \space z \neq 2 $